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Strugling with one probability question, could use some help

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The question goes as follows: In a certain town three newspapers are published. 20% of the population read A, 16% read B, 14% read C, 8& read A and B, 5% read A and C, 4% read B and C, and 2% read all 3. A person is selected at random. Use a venn diagram to help determine the probability that the person reads:

a.) none of the papers b.) at least one of the papers c.) exactly one of the papers d.) either A or B e.) A given that the person reads at least 1 paper f.) C, given that the person reads either A or B or both.

Now I've tried to solve this problem for about an hour now, cant seem to even get the correct answer for a...

answers are: a.) 13/20 b.) 7/20 c.) 11/50 d.) 7/25 e.) 4/7 f.) 1/4

gah got my november math sl retakes in 2 weeks and I'm stuck on this... I regret not taking math studies, but oh well if anyone does help thank you !

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To reduce the confusion of percentages, let's assume the town (universal set) has 100 people.

  1. That means the circle for A=20 people, B=16 people, and C=14 people.
  2. We will then skip ahead and see "2% read all". That means 2 people are at AnBnC (center).
  3. Now 8 people read A and B, and we know 2 read all. Then 6 people read A and B but not C.
  4. 5 people read A and C, and we know 2 read all. Then 3 read A and C but not B.
  5. Similarly, 2 read B and C but not A.
  6. The remaining numbers in circles are filled in such that, all the numbers in A add up to 20, B up to 16, and C up to 14.
  7. The universal set is 100 but the sum of all number in circles is 35. Therefore there's 65 outside.

I just did a venn diagram to illustrate what I just described.

post-24693-010921400 1287522054_thumb.pn

a) The number outside the circles represent the number of people who don't read. Therefore Probability = 65/100 = 13/20

b) The total number inside circles = 35, this is the number of people who read at least one newspaper (A, B, or C). Probability = 35/100 = 7/20

c) Sum of numbers in no intersection = 9+6+7 = 22, number of people who read just one (hence not being in intersection). Probability = 22/100 = 11/50

d) It's better to say A or B. Adding 'Either' implies (A or B but not both) which isn't what was meant here judging from the answer. Sum of all number in A or B, 9+3+6+2+6+2=28, number of people reading A or B. Probability = 28/100 = 7/25

e) Number of people reading A = 20 (given in question) and 35 read at least one newspaper (from part b.) Probability = 20/35 = 4/7

f) Again it would have been better to just say 'A or B' instead of adding 'either' and having to avoid confusion by adding 'or both'. 'Either A or B or both' is the same as just 'A or B'. Anyway, number of people reading A or B = 28 (from d.), and of these, sum of numbers in C too = 3+2+2 = 7, number of people reading C given that they read A or B. Probability = 7/28 = 1/4.

Sorry, can't give better explanation through text; ask someone to explain it to you in person if you still don't understand.

Edited by Gene-Peer
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you have no idea how grateful I am ! thank :( and nah I understand now, I see what I did wrong. I didnt subtract the amount of people who read all newspapers from the amount of people who read A and B and B and C.

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Kind regards,

IB Survival Staff

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